Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $y = \dfrac{a^2 + 9a}{-a^2 + 2a + 3} \times \dfrac{2a^2 + 10a + 8}{-a^2 - 4a} $
Answer: First factor out any common factors. $y = \dfrac{a(a + 9)}{-(a^2 - 2a - 3)} \times \dfrac{2(a^2 + 5a + 4)}{-a(a + 4)} $ Then factor the quadratic expressions. $y = \dfrac {a(a + 9)} {-(a + 1)(a - 3)} \times \dfrac {2(a + 1)(a + 4)} {-a(a + 4)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac {a(a + 9) \times 2(a + 1)(a + 4) } { -(a + 1)(a - 3) \times -a(a + 4)} $ $y = \dfrac {2a(a + 1)(a + 4)(a + 9)} {a(a + 1)(a - 3)(a + 4)} $ Notice that $(a + 1)$ and $(a + 4)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac {2a\cancel{(a + 1)}(a + 4)(a + 9)} {a\cancel{(a + 1)}(a - 3)(a + 4)} $ We are dividing by $a + 1$ , so $a + 1 \neq 0$ Therefore, $a \neq -1$ $y = \dfrac {2a\cancel{(a + 1)}\cancel{(a + 4)}(a + 9)} {a\cancel{(a + 1)}(a - 3)\cancel{(a + 4)}} $ We are dividing by $a + 4$ , so $a + 4 \neq 0$ Therefore, $a \neq -4$ $y = \dfrac {2a(a + 9)} {a(a - 3)} $ $ y = \dfrac{2(a + 9)}{a - 3}; a \neq -1; a \neq -4 $